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问题描述
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
Calling next() will return the next smallest number in the BST.
Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
解决思路
中序遍历+非递归（辅助栈）
时间/空间复杂度均为O(logn).
程序
public class BSTIterator {
private Stack<TreeNode> s;
public BSTIterator(TreeNode root) {
s = new Stack<>();
pushLeft(root);
}
private void pushLeft(TreeNode node) {
TreeNode cur = node;
while (cur != null) {
s.push(cur);
cur = cur.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !s.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode node = s.pop();
pushLeft(node.right);
return node.val;
}
}
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原文地址：http://www.cnblogs.com/harrygogo/p/4655390.html
